54. Spiral Matrix

Q: Why is the four-boundaries version preferred?

It uses O(1) extra space and the loop structure makes the spiral pattern explicit — anyone reading the code can see the four sides. The visited-matrix version is correct but less expressive.

Q: What happens with a single row or column?

The two inner `if` checks handle this. For a 1xN matrix, after the top-row pass `top` becomes 1 and `top > bottom` halts the inner branches — preventing double-counting.

mediumAsked at Netflix

Given an m x n matrix, return all elements in spiral order. Netflix asks this to test whether you can manage four-boundary state without going out of bounds — the canonical 'index gymnastics' question.

By Alex Chen, Founder, InterviewChamp.AI · Last verified 2026-05-19

Source citations

Public interview reports confirming this problem appears in Netflix loops.

Glassdoor (2026-Q1)— Netflix L4 onsite reports cite this as the iterative-matrix problem on the algorithms screen.
Blind (2025-10)— Netflix SWE writeups describe this as the bounds-management warmup before a harder follow-up.

Problem

Given an m x n matrix, return all elements of the matrix in spiral order (clockwise, starting from the top-left).

Constraints

m == matrix.length
n == matrix[i].length
1 <= m, n <= 10
-100 <= matrix[i][j] <= 100

Examples

Example 1

Input

matrix = [[1,2,3],[4,5,6],[7,8,9]]

Output

[1,2,3,6,9,8,7,4,5]

Example 2

Input

matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]

Output

[1,2,3,4,8,12,11,10,9,5,6,7]

Approaches

1. Direction array + visited matrix

Use an array of (dr, dc) for the four directions. Walk step by step; if next cell is out of bounds or visited, rotate direction.

Time: O(m * n)
Space: O(m * n) for visited

function spiralOrderVisited(matrix) {
  const m = matrix.length, n = matrix[0].length;
  const out = [];
  const seen = Array.from({ length: m }, () => new Array(n).fill(false));
  const dirs = [[0, 1], [1, 0], [0, -1], [-1, 0]];
  let r = 0, c = 0, d = 0;
  for (let i = 0; i < m * n; i++) {
    out.push(matrix[r][c]);
    seen[r][c] = true;
    const nr = r + dirs[d][0];
    const nc = c + dirs[d][1];
    if (nr < 0 || nr >= m || nc < 0 || nc >= n || seen[nr][nc]) d = (d + 1) % 4;
    r += dirs[d][0]; c += dirs[d][1];
  }
  return out;
}

Tradeoff: Easiest to reason about. The visited matrix uses O(m * n) extra space which is acceptable here but rejected by senior interviewers who want O(1) extra.

2. Four boundaries (optimal)

Maintain top, bottom, left, right boundaries. Walk top row left-to-right, then right column top-to-bottom, then bottom row right-to-left (if rows remain), then left column bottom-to-top (if cols remain). Shrink each boundary after use.

Time: O(m * n)
Space: O(1) extra (excluding output)

function spiralOrder(matrix) {
  const out = [];
  let top = 0, bottom = matrix.length - 1;
  let left = 0, right = matrix[0].length - 1;
  while (top <= bottom && left <= right) {
    for (let c = left; c <= right; c++) out.push(matrix[top][c]);
    top++;
    for (let r = top; r <= bottom; r++) out.push(matrix[r][right]);
    right--;
    if (top <= bottom) {
      for (let c = right; c >= left; c--) out.push(matrix[bottom][c]);
      bottom--;
    }
    if (left <= right) {
      for (let r = bottom; r >= top; r--) out.push(matrix[r][left]);
      left++;
    }
  }
  return out;
}

Tradeoff: O(1) extra space. The two `if` checks after the second pair of for-loops are critical — they prevent re-visiting a row/column when the spiral has narrowed to a single line.

Netflix-specific tips

Netflix interviewers grade two specific things: (1) Do you understand WHY the two `if` checks (top <= bottom and left <= right) are needed in the boundaries version? They prevent re-traversing the last row or column. (2) Can you trace through a 1xN or Nx1 matrix in your head? Most candidates' first version fails one of those edge cases.

Common mistakes

Forgetting the inner `if (top <= bottom)` check before the right-to-left pass — overwrites the same row in a 1xN matrix.
Using a single visited matrix when the boundaries approach is cleaner — works but signals you didn't see the better solution.
Off-by-one on the inner for-loop bounds (`c < right` vs `c <= right`).

Follow-up questions

An interviewer at Netflix may pivot to one of these next:

Spiral Matrix II (LC 59) — generate an n x n matrix filled in spiral order.
Spiral Matrix III (LC 885) — start at (r, c) and walk a spiral covering an R x C grid.
Rotate Image (LC 48) — related index-gymnastics warmup.

Solve it now

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Output

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FAQ

Why is the four-boundaries version preferred?

It uses O(1) extra space and the loop structure makes the spiral pattern explicit — anyone reading the code can see the four sides. The visited-matrix version is correct but less expressive.

What happens with a single row or column?

The two inner `if` checks handle this. For a 1xN matrix, after the top-row pass `top` becomes 1 and `top > bottom` halts the inner branches — preventing double-counting.

Free learning resources

Curated free links for this problem.

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